Question

In: Chemistry

Physical Chemistry A)The electrolytic conductivity of 0.10 M solution of acetic acid is 5.3x 10-5 Scm-1....

Physical Chemistry

A)The electrolytic conductivity of 0.10 M solution of acetic acid
is 5.3x 10-5 Scm-1. The cell constant was 0.10 cm-1. Neglecting activity
coefficient, Calculate the acid dissociation constant (Ka).
(Λmo for acetic acid = 390. S cm2 mol-1)

B)The observed osmotic pressure of 0.1 M acetic acid solution is
2.45 atm at 25.0 oC. Calculate the Van’t Hoff factor (i) and the degree of
ionization (α). Compare your value with that obtained from problem 1.
Comment on your answer.

Expert Solution

A). electrolytic conductivity, k = 5.3x 10-5 Scm-1

concentration, c = 0.10 M

cell constant = 0.10 cm-1

Λmo for acetic acid = 390. S cm2 mol-1

Λm = Λmo - k*

= 390 - 5.3 x 10^-5 *

= 390 - 1.676 x 10^-5

= 389.99

degree of dissociation, = Λm / Λmo

= 389.99 / 390

= 0.99

Acid dissociation constant, Ka = *c / (1 - )

= 0.99² * 0.10 / (1 - 0.99)

= 0.09801 / 0.01

= 9.801

Thus, acid dissociation constant, Ka = 9.801

B). Osmotic pressure, = 2.45 atm

concentration, c = 0.1 M

Temperature, T = 25 oC = 298 K

We know that -

= i *c*R*T

2.45 = i*0.1*0.0821*298

2.45 = i*2.446

i = 2.45 / 2.446

i = 1.0

Van't Hoff factor, i = 1.0

Acetic acid ionize as -

CH3COOH CH3COO- + H+

Acetic acid ionize into 2 ions.

n = 2

1 = 1 + (2 - 1)

= 0

degree of ionization (α) = 0

queen_honey_blossom answered 1 month ago

In the solution containing both 0.10 M acetic acid and 0.10 M sodium acetate, the acetic...

In the solution containing both 0.10 M acetic acid and 0.10 M sodium acetate, the acetic acid undergoes ionization. The chemical equation for this ionization reaction is the same as for a solution containing acetic acid alone. The difference is that the initial concentration of acetate ion (before any ionization reaction occurs) for the solution containing acetic acid alone is zero, whereas the initial concentration of acetate ion is 0.10 M in your solution containing both acetic acid and sodium...

1.) Consider a solution that is 0.10 M Acetic Acid (HC2H3O2) and 0.10 M Sodium Acetate (NaC2H3O2).

1.) Consider a solution that is 0.10 M Acetic Acid (HC2H3O2) and 0.10 M Sodium Acetate (NaC2H3O2).HC2H3O2 + H2O <---> C2H3O2- + H3O+a.) What happens if you add HCl (acid)?b.) What happens if you add NaOH (base)?2.) NH3 + H2O <---> NH4+ + OH-If you add HCl, which way will it shift?

imagine mixing 25.0 ml of 0.10 M acetic acid (Ka=1.8*10^-5) with 25.0 ml of 0.10 M...

imagine mixing 25.0 ml of 0.10 M acetic acid (Ka=1.8*10^-5) with 25.0 ml of 0.10 M sodium acetate and determine a) The initial pH of the buffer b) pH of 20.0 ml sample of buffer with 1.0 ml of 0.10M Hal added c) pH of another 20 mL sample of buffer with 1.0 ml of 0.10 M MaOH added

a 25.0 mL solution if 0.10 M acetic acid is titrated with 0.1 M NaOH solution....

a 25.0 mL solution if 0.10 M acetic acid is titrated with 0.1 M NaOH solution. the pH equivalence is

Compare the calculated percent ionization of acetic acid in the solution containing both 0.10 M acetic...

Compare the calculated percent ionization of acetic acid in the solution containing both 0.10 M acetic acid and 0.10 M sodium acetate to the calculated percent ionization for the solution containing 0.10 M acetic acid alone. In which solution is the percent ionization the lowest? Clearly and completely explain why this occurs.

1. A buffer solution consists of 0.25 M acetic acid (CH3COOH) and 0.10 M sodium acetate...

1. A buffer solution consists of 0.25 M acetic acid (CH3COOH) and 0.10 M sodium acetate (CH3COONa). The Ka of acetic acid is 1.8×10−5. a. Calculate the equilibrium pH using the equilibrium mass action expression. b. Calculate the equilibrium pH using the Henderson-Hasselbalch equation approach. 2. Consider a buffer solution that consists of both benzoic acid (C6H5COOH) and sodium benzoate (C6H5COONa) and has a pH of 4.18. The concentration of benzoic acid is 0.40 M. The pKa of benzoic acid...

A buffer solution that is 0.10 M sodium acetate and 0.20 M acetic acid is prepared....

A buffer solution that is 0.10 M sodium acetate and 0.20 M acetic acid is prepared. Calculate the initial pH of this solution. The Ka for CH3COOH is 1.8 x 10-5 M. As usual, report pH to 2 decimal places. A buffered solution resists a change in pH. Calculate the pH when 21.1 mL of 0.031 M HCl is added to 100.0 mL of the above buffer.

Calculate the percent ionization and the expected intial pH for the 0.10 M acetic acid solution

For the titration of 75 mL of 0.10 M acetic acid with 0.10 M NaOH, calculate...

For the titration of 75 mL of 0.10 M acetic acid with 0.10 M NaOH, calculate the pH. For acetic acid, HC2H3O2, Ka = 1.8 x 10-5. (a) before the addition of any NaOH solution. (b) after 25 mL of the base has been added. (c) after half of the HC2H3O2 has been neutralized. (d) at the equivalence point.

Calculate the pH of 0.10 M sodium acetate solution ( Ka, acetic acid = 1.76 x...

Calculate the pH of 0.10 M sodium acetate solution ( Ka, acetic acid = 1.76 x 10-5 ).